Determine the reaction on the rigid body subjected to the distributed load shown.

T=138 °C

Explanation:

Given that

m = 0.028 kg

Net work output W= 60 KJ

T₂=2T₁

As we know that efficiency of Carnot heat engine given as

η = 0.5

We know that

Qa=heat addition

W= net work output

Qa= 120 KJ

From first law

Qa= W+ Qr

Qr= 120 - 60

Qr= 60 KJ  

Qr Is the heat rejection.

Heat rejection per unit mass

Qr=60 / 0.028 = 2142.85 KJ/kg

Qr= 2142.85 KJ/kg

Temperature at which latent heat of steam is  2142.85 KJ/kg will be our answer.

T=138 °C

The temperature corresponding to 2142.85 KJ/kg will be 138 °C.

T=138 °C